An aqueous solution containing 15.62 g of nickel(II) nitrate is added to an aqueous solution containing 11.51 g of sodium phosphate. A precipitate forms.
1) Predict the products (write the formula).
2) Write a balanced chemical equation.
3) Determine the limiting reactant (LR).
4) Calculate the mass of each product and the reactant in excess (XS).
3Ni(NO3)2 (aq) + 2Na3PO4 (aq) --------> Ni3(PO4)2 (s) + 6NaNO3 (aq)
|
Compound |
3Ni(NO3)2 |
2Na3PO4 |
Ni3(PO4)2 |
6NaNO3 |
|
Mass (g), initial |
15.62 |
11.51 |
0 |
0 |
|
MM (g/mol) |
182.2703 |
163.9407 |
366.0229 |
84.9947 |
|
Moles, initial |
0.08570 |
0.07021 |
0 |
0 |
|
Change in moles |
-3x |
-2x |
+x |
+6x |
|
Moles, final |
0 |
0.01308 |
0.02857 |
0.1714 |
|
Mass (g), final |
0 |
2.144 |
10.46 |
14.57 |
x = 0.028566
Find the L.R., mass of XS reactant, and mass of N2, CO2, and H2O in the following reaction.
C2H8N2 (g) + 2N2O4 (g) --------> 3N2 (g) + 2CO2 (g) + 4H2O (g)
0.900 g 2.20 g ?g ?g ?g
|
Compound |
C2H8N2 |
2N2O4 |
3N2 |
2CO2 |
4H2O |
|
Mass (g), initial |
0.900 |
2.20 |
0 |
0 |
0 |
|
MM (g/mol) |
60.10 |
92.012 |
28.014 |
44.01 |
18.0154 |
|
Moles, initial |
0.014975 |
0.023910 |
|
? |
? |
|
Change in moles |
-x |
-2x |
+3x |
+2x |
+4x |
|
Moles, final |
0.003020 |
0 |
0.035865 |
0.02391 |
0.04782 |
|
Mass (g), final |
0.1815 |
0 |
1.0047 |
1.0523 |
0.86150 |
x = 0.011955